Problem: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $23$ years; the standard deviation is $5.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living longer than $28.6$ years.
Solution: $23$ $17.4$ $28.6$ $11.8$ $34.2$ $6.2$ $39.8$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $23$ years. We know the standard deviation is $5.6$ years, so one standard deviation below the mean is $17.4$ years and one standard deviation above the mean is $28.6$ years. Two standard deviations below the mean is $11.8$ years and two standard deviations above the mean is $34.2$ years. Three standard deviations below the mean is $6.2$ years and three standard deviations above the mean is $39.8$ years. We are interested in the probability of a porcupine living longer than $28.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the porcupines will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $17.4$ years and the other half $({16\%})$ will live longer than $28.6$ years. The probability of a particular porcupine living longer than $28.6$ years is ${16\%}$.